Two Plus Two Newer Archives  

Go Back   Two Plus Two Newer Archives > Other Topics > Science, Math, and Philosophy
FAQ Community Calendar Today's Posts Search

Reply
 
Thread Tools Display Modes
  #21  
Old 06-22-2007, 12:22 AM
borisp borisp is offline
Senior Member
 
Join Date: Nov 2004
Posts: 201
Default Re: The Nash Equilibrium and the traveller\'s dilemma

[ QUOTE ]
[ QUOTE ]
The confusion appears to be that the dilemma assumes (while stating it does not) that the travellers are playing against each other where in the story they are not playing a game but just both trying to get as much money as possible.

[/ QUOTE ]

The solution simply assumes that each player tries to make the most money possible.

[/ QUOTE ]
This is the truth, but not quite the "whole truth." The whole truth would add that each player KNOWS that their opponent is perfectly rational, is seeking the same goal, and that their opponent knows this of them. Then the solution follows.

The marginal utility of a guaranteed 2$ pales in comparison to the marginal utility of a possible 100$. This is why "normal" people would choose 100, and why they would expect other normal people to do so. A rational agent can "beat" this by playing 99, but has gained only 1$ relative to the "stable" strategy of both players playing 100$. In real life, we make these decisions as if we plan to play the game over and over, and a player who consistently played 99, in an effort to get this tiny edge, would most likely not get invited to play this positive sum game very often.

Hence we see how it relates to poker: the "reasonable" choice is the one that keeps the game going and is profitable for everybody. The "rational" choice is one that is unexploitable, but it also ensures no one will want to play against you.
Reply With Quote
  #22  
Old 06-22-2007, 12:44 AM
CallMeIshmael CallMeIshmael is offline
Senior Member
 
Join Date: Dec 2004
Location: Tis the season, imo
Posts: 7,849
Default Re: The Nash Equilibrium and the traveller\'s dilemma

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
The confusion appears to be that the dilemma assumes (while stating it does not) that the travellers are playing against each other where in the story they are not playing a game but just both trying to get as much money as possible.

[/ QUOTE ]

The solution simply assumes that each player tries to make the most money possible.

[/ QUOTE ]
This is the truth, but not quite the "whole truth." The whole truth would add that each player KNOWS that their opponent is perfectly rational, is seeking the same goal, and that their opponent knows this of them. Then the solution follows.

[/ QUOTE ]

yeah, very true


Im not sure if you read the linked thread, but, in it, a lot of people made the mistake "the solution assumes they are trying to beat each other", when the solution most certainly does not assume that, and I just wanted to point that out.
Reply With Quote
  #23  
Old 06-22-2007, 12:55 AM
borisp borisp is offline
Senior Member
 
Join Date: Nov 2004
Posts: 201
Default Re: The Nash Equilibrium and the traveller\'s dilemma

[ QUOTE ]
...Im not sure if you read the linked thread...

[/ QUOTE ]
I stopped after about 4 pages, to stifle the indigestion [img]/images/graemlins/smile.gif[/img]
Reply With Quote
  #24  
Old 06-22-2007, 04:18 AM
iggymcfly iggymcfly is offline
Senior Member
 
Join Date: Jun 2006
Posts: 3,784
Default Re: The Nash Equilibrium and the traveller\'s dilemma

I think this is kind of stupid. Why is it "rational" to assume that your opponent is going to try to use 98th level thinking on you? And to risk a potential $99 to make $2 when he does?

It seems to me like the maximal solution is to assume that a lot of people will play 100 and pick 99 which maximizes your expectation most often in the real world while the optimal unexploitable solution would be to assume that your opponent is equally likely to pick any number and pick either 96 or 97 since those numbers have the highest sum totals against all possibilities.

Picking 2 is stupid since it only does best against 2 and 3 and is the worst choice you can pick against any other number.
Reply With Quote
  #25  
Old 06-22-2007, 04:26 AM
borisp borisp is offline
Senior Member
 
Join Date: Nov 2004
Posts: 201
Default Re: The Nash Equilibrium and the traveller\'s dilemma

[ QUOTE ]
I think this is kind of stupid. Why is it "rational" to assume that your opponent is going to try to use 98th level thinking on you? And to risk a potential $99 to make $2 when he does?

It seems to me like the maximal solution is to assume that a lot of people will play 100 and pick 99 which maximizes your expectation most often in the real world while the optimal unexploitable solution would be to assume that your opponent is equally likely to pick any number and pick either 96 or 97 since those numbers have the highest sum totals against all possibilities.

Picking 2 is stupid since it only does best against 2 and 3 and is the worst choice you can pick against any other number.

[/ QUOTE ]
Before you start reasoning, think about whether the opposite side will hear your reasoning.
Reply With Quote
  #26  
Old 06-22-2007, 06:00 AM
Galwegian Galwegian is offline
Senior Member
 
Join Date: Jul 2006
Posts: 281
Default Re: The Nash Equilibrium and the traveller\'s dilemma

[ QUOTE ]
I suck with Google, but typing in

definition "maximal strategy" "game theory"

gives hits from Cornell and Jstor at the top. Pick your poison.

In my limited experience most people use the term optimal when they are really thinking about a maximal strategy and most people don't actually know what an optimal strategy is. It's usually not the correct strategy to play in games with opponents that are casual gamers.

[/ QUOTE ]
OK - so according to the University of Alberta game theory pages, a maximal strategy is what poker players normally refer to as an exploitive strategy (I suppose I should say maximally exploitive strategy). I definitely understand the difference between these concepts.

[ QUOTE ]
You can also understand that the idea that if we knew the second player's number, the optimal & maximal strategies are known and agree (e.g if their number is 100, ours is 99).

[/ QUOTE ] This doesn't make any sense to me. The term optimal specifically refers to the situation where you assume that the opponent plays perfectly i.e. inexploitably. If you know that opponent's number is 100 (for example), then it makes no sense to talk about optimal strategy, hence it makes no sense to wonder whether it coincides with maximal strategy.
Reply With Quote
  #27  
Old 06-22-2007, 06:09 AM
Galwegian Galwegian is offline
Senior Member
 
Join Date: Jul 2006
Posts: 281
Default Re: The Nash Equilibrium and the traveller\'s dilemma

[ QUOTE ]
Off the top of my head, Id say you might have problems finding a very good poker equivalent to the situation in the TD, since poker is zero sum and the TD isnt, AND the weird results of the TD are specifically because of the fact that it isnt zero sum.

[/ QUOTE ]
Two observations. Tournament poker is not zero sum. There are many situations (for example) where there are just two players left in a hand, and where their strategies will result in changes of equity for the players not in the hand. Easy example - say there are 4 players left in a standard STT (3 places paid) and it is folded to the small blind. Now we have a heads up between the blinds that is not a zero sum game. In fact, virtually every tournament situation where some players have folded their hands results in a non zero sum game between the remaining players.

Second, it is not clear to me at all that the non zero sum nature of TD has anything to do with the surprising Nash equilibrium
Reply With Quote
  #28  
Old 06-22-2007, 08:28 AM
Piers Piers is offline
Senior Member
 
Join Date: Sep 2002
Posts: 1,616
Default Re: The Nash Equilibrium and the traveller\'s dilemma

[ QUOTE ]
This is the truth, but not quite the "whole truth." The whole truth would add that each player KNOWS that their opponent is perfectly rational, is seeking the same goal, and that their opponent knows this of them. Then the solution follows.


[/ QUOTE ]

So what does perfectly rational mean here? I am fairly sure no human is ‘perfectly rational’, so how can a perfectly rational entity assume a human opponent is perfectly rational?

I believe in any real situation that the proportion of people that would choose $2 is so small that it would be bad judgement to assume that your opponent is one of them. Its possible one of the individuals is a game theory geek who thinks looking clever to himself is what’s important and is not really thinking about the other person, and the other person knows it (But even so I would choose $99 or $100 just in case I am wrong)..

Of course if units were billions of dollars not dollars, then I would choose $2billion, but that’s something else.

Approaching this only as a mathematical toy, the fact that it talks about rational humans when no such entity exists is fine. However when you start sounding like you are talking about real people in the real world you cannot be so blasé about the existence of these curiously defined rational humans.
Reply With Quote
  #29  
Old 06-22-2007, 02:42 PM
CallMeIshmael CallMeIshmael is offline
Senior Member
 
Join Date: Dec 2004
Location: Tis the season, imo
Posts: 7,849
Default Re: The Nash Equilibrium and the traveller\'s dilemma

[ QUOTE ]
the optimal unexploitable solution would be to assume that your opponent is equally likely to pick any number and pick either 96 or 97 since those numbers have the highest sum totals against all possibilities.

[/ QUOTE ]

Do you know what "unexploitbale" means in the GT sense?

Because neither of those choices are
Reply With Quote
  #30  
Old 06-22-2007, 02:52 PM
wazz wazz is offline
Senior Member
 
Join Date: Jun 2006
Location: London
Posts: 2,560
Default Re: The Nash Equilibrium and the traveller\'s dilemma

I hate to point this out, but if the objective is to 'make as much money as possible' you should choose $99 or $100. The solution can only be $2 if the objective is to make more money than the other person.

On further thinking, $2 couldn't be the right answer either - assuming the other guy picks $2 as well, which is a 'safe' assumption given all this reasoning, you can make $3 by choosing $1. EDIT oh or maybe not.
Reply With Quote
Reply


Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -4. The time now is 12:15 AM.


Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2024, vBulletin Solutions Inc.